Section C Section C Checkpoint

Problem 1

Select all the expressions that can give the value of 6÷326 \div \frac{3}{2}.
A
6326 \boldcdot \frac{3}{2}
B
(62)÷3(6 \boldcdot 2) \div 3
C
(6÷12)÷3(6 \div \frac{1}{2}) \div 3
D
62136 \boldcdot 2 \boldcdot \frac{1}{3}
E
(6÷2)÷3(6 \div 2) \div 3
F
6236 \boldcdot \frac{2}{3}
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Solution
B, C, D, F
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Sample Response
B, C, D, F

Problem 2

Calculate each quotient. Show your reasoning.

  1. 98÷32\frac{9}{8} \div \frac{3}{2}

  2. 2110÷152 \frac{1}{10} \div \frac{1}{5}
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Solution
  1. 34\frac{3}{4} (or equivalent). Sample reasoning:
    • 98÷32=9823=1824=34\frac{9}{8} \div \frac{3}{2} = \frac{9}{8} \boldcdot \frac{2}{3} = \frac{18}{24} = \frac{3}{4}
    • The quotient is the unknown factor in the multiplication equation 32?=98\frac{3}{2} \boldcdot {?} = \frac{9}{8}. That number is 34\frac{3}{4}.
  2. 212\frac{21}{2} or 101210\frac{1}{2} (or equivalent). Sample reasoning:
    • 2110÷15=21105=10510=10510\frac{21}{10} \div \frac{1}{5} = \frac{21}{10} \boldcdot 5 = \frac{105}{10} = 10\frac{5}{10}
    • There are 10 groups of 15\frac{1}{5} in 2 wholes, and 110\frac{1}{10} is 12\frac{1}{2} a group of 15\frac{1}{5}, so there are 101210\frac{1}{2} groups of 15\frac{1}{5} in 21102\frac{1}{10}.
Show Sample Response
Sample Response
  1. 34\frac{3}{4} (or equivalent). Sample reasoning:
    • 98÷32=9823=1824=34\frac{9}{8} \div \frac{3}{2} = \frac{9}{8} \boldcdot \frac{2}{3} = \frac{18}{24} = \frac{3}{4}
    • The quotient is the unknown factor in the multiplication equation 32?=98\frac{3}{2} \boldcdot {?} = \frac{9}{8}. That number is 34\frac{3}{4}.
  2. 212\frac{21}{2} or 101210\frac{1}{2} (or equivalent). Sample reasoning:
    • 2110÷15=21105=10510=10510\frac{21}{10} \div \frac{1}{5} = \frac{21}{10} \boldcdot 5 = \frac{105}{10} = 10\frac{5}{10}
    • There are 10 groups of 15\frac{1}{5} in 2 wholes, and 110\frac{1}{10} is 12\frac{1}{2} a group of 15\frac{1}{5}, so there are 101210\frac{1}{2} groups of 15\frac{1}{5} in 21102\frac{1}{10}.